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Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.

$$x=2 \sqrt{y}, x=0, y=9 ; \quad \text { about the } y-axis$$

$$

162 \pi

$$

Applications of Integration

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you given X equals two squared of what I X equals zero y equals nine and we're going around the y axis as we're graphing. This you may not be sure had a graph X equals two squared of Why So here's a couple options. Obviously you could pick numbers to fill in for Why? Like if I let y equal one, then X equals two. So that ordered pair the Y values one. The Xlu is too. Okay, you go point white point like that. But probably the easier way is to write it in the function form like you're used. Teoh. So I would just rewrite this in terms of why equals for a second, because on this want to have the luxury that that actually works doesn't always work. So maybe I just make a table here. If I square both sides to get rid of the square root, I would have Why equals X squared over four. That obviously makes a really fat parabola, right? That's a 1/4 x squared. But if you look back up at the original, the Onley kind of answers you can get for Accer positive. So even though what I just solved Makes a wide parabola. I can't use the left half of it because up in my original function think about it. If you plugged in negative numbers, you can't have a square root of a negative. You're only getting positive outputs there for X. So this left part of the graph actually doesn't exist for your original. So helpful to rewrite it. But beware if your domain is restricted. Now, the good news is your boundaries helped you because the boundaries otherwise or X equals zero, which is your y axis, and why equals nine. And I'm just gonna go ahead, put this here. We'll say we're really zoomed out right now. Eventually, that's how they would meet. So the area we're bounding is this entire section shaded in green. We're revolving around the y axis. So as I consider that shape noticed that all of it comes right up alongside that. Why access? There's no space is nothing missing, which means that I will be able to use a disk. So the volume for the disk is pie in a girl R squared in terms of whatever access your revolving around. So I have d y here because I'm going around the Y axis, and that would mean I need the lowest and the highest y values. So for this shape, what's the lowest possible? Why value wouldn't have to be here at zero, and then the highest possible why value is that boundary of Given, which happens at nine. So the only other thing I need to be able to fill in to solve this is that r squared Now R squared is the radius, and the radius on this one goes from the center to the edge of our figure, the centers where I'm rotating Thea edge of the figure is defined by that proble shape. So, really, for our in this case, in terms of why we're thinking about right minus left. So the right function is that X equals two square root of why the left function is X equals zero. Now, how did I know to use two squared of why and not the proble? Easy? Because I'm going around the y axis. I have d Y as my variable for integration. And if I'm using d y, then I have to use why and my function as well. So we're gonna fill in two squared of why into the integral. Now this one is pretty easy to go through and square both terms we get for y. If we integrate this for, why then we would have to. I squared. We raised the power raise that degree up one divide by that new exponents. We get two y squared and then we have to evaluate for the upper and the lower. So we plug in nine and then we plug in zero. Obviously that zero term drops out. So 162 pies my final answer for this volume.

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Applications of Integration